Question

What is the difference between the right-hand limit and the left-hand limit of
f(x) =|x − 3|/x − 3 as x approaches 3?

no difference
1 unit
2 units
3 units

Answer 1

From the left side of x=3,\[\Large\rm |x-3|=-(x-3)\]Plug that in for your left-sided limit.


From the right side of x=3,\[\Large\rm |x-3|=x-3\]Plug that in for the right-sided limit :)

Answer 2

i still dont understand. sorry i new to this

Answer 3

\[\Large\rm |x-3|=\cases{x-3,&x>3\\ \Large\rm -(x-3), &x<3}\]
\[\Large\rm \lim_{x\to3^-} \frac{|x-3|}{x-3}=\lim_{x\to3^-} \frac{-(x-3)}{x-3}\]The absolute function is negative when we're left of 3.
Simplify that to get your left-sided limit.

Answer 4

\[\Large\rm \lim_{x\to3^+}\frac{|x-3|}{x-3}=\lim_{x\to3^+}\frac{x-3}{x-3}\]|x-3| is positive when we're on the right side of 3.
Simplify! :)

What do you get for your two limits?

Answer 5

\(\bf lim_{x\to 3}\ \cfrac{|x-3|}{x-3}\implies
\begin{cases}
lim_{x\to {\color{red}{ 3^-}}}\ \cfrac{-(x-3)}{x-3}
\\ \quad \\
\bf lim_{x\to {\color{red}{ 3^+}}}\ \cfrac{+(x-3)}{x-3}
\end{cases}\)

Answer 6

i think its no difference right?

Answer 7

@davidza6496 I thought @kirbykirby did this with you earlier?

Answer 8

Try looking at the graph of \(\large y=\frac{|x-3|}{x-3} \) and compare the y-values both to the left and right of x=3.

Answer 9

is the difference 2 i think i got

Answer 10

So you got -1 from the left side, and 1 from the right side?
distance of 2 between them?
Yes good job! :)